JEE Main 2015 — Gravitation Question with Solution

Central idea is, consider the cavity as negative a mass and apply the superposition of gravitational potential.

Consider the cavity formed in a solid sphere as shown in the figure.

The gravitational potential at infinite distance,

$V=\frac{-GM}{r}$

$\Rightarrow V\left(\infty \right)=0$

According to the question, we can write potential at an internal point $P$ due to complete solid sphere,

$V_s=-\frac{GM}{2R^3}\left[3R^2-{\left(\frac{R}{2}\right)}^2\right]$

$=\frac{-GM}{2R^3}\left[3R^2-\frac{R^2}{4}\right]$

$=\frac{-GM}{2R^3}\left[\frac{11R^2}{4}\right]=\frac{-11GM}{8R}$

Mass of removed part

$=\frac{M}{\frac{4}{3}\times \pi R^3}\times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3=\frac{M}{8}$

The potential at a point $P$ due to removed part

$V_C=\frac{-3}{2}\times \frac{\left({\displaystyle \frac{GM}{8}}\right)}{R}=\frac{-3GM}{8R}$

Thus, potential due to the remaining part at the point $P\text{,}$

$V_P=V_s-V_c=\frac{-11GM}{8R}-\left(-\frac{3GM}{8R}\right)$

$\frac{\left(-11+3\right)GM}{8R}=\frac{-GM}{R}$